This notes explain how to construct Luroth quartic from a vector bundle with invariants (r=2,c1=0,c2=4) on P^2. Luroth quartic is a plane quartic curve circumscribed about a 5-side on the plane. Thus we get a map from M(2,0,4) on P^2 to the variety of Luroth's quartics. Step 1. Let E be an element in M(4)=M(2,0,4) on P^2. We have chi(E(1)) = 2, and h^2(E(1)) = 0 by stability. (In general, H^2(E(i)) = 0 for i >= -3, E in M(2,0,n)). It follows that h^0(E(1)) >= 2. In fact, one can prove that generic E has h^0(E(1)) = 2. Step 2. Consider the variety M~ of pairs (E,V), where V is a 2-dim. vector subspace in H^0(E(1)). Then M~ is barational to M, and a generic point in M~ gives an exact sequence 0 -> V tensor O -> E(1) -> B -> 0, where B is a line bundle on a plane conic C isomorphic to O(-1). (Note that E(1) has Chern classes (2,2,5)). Step 3. Let's choose a section s in V < H^0(E(1)). Let Z(s) be the cycle of zeroes of s. It has degree c_2(E(1)) = 5. We have Z(s) < C, since the map V O -> E(1), clearly, degenerates at the points of Z(s). Let p1, p2 be 2 points in Z(s), and $l$ be a line through p1,p2. Lemma: $l$ is a jumping line for E, i.e., E|_l is not isomorphic to O+O. (Proof in the Appendix). Step 4. Thus we get . a conic C in the projective plane P; . 5 points x_i on C; . 10 lines l_ij through x_i and x_j such that l_ij is a jumping line for E. Barth proves that jumping lines for E for a curve S of degree 4 in the dual projective plane P^2. In the dual picture we get . a conic C^dual in the projective plane P^dual; . 5 tangent lines X_i to C^dual; . 10 intersection points L_ij of X_i and X_j such that . L_ij is on the quartic S. The configuration (C^dual, L_ij) is called Luroth's quartic. ------------------------------------------------------------ Appendix. Let s be a section of E(1). it gives an exact sequence s 0 -> O -> E(1) -> J_z(2) -> 0 (a) Let l be a point not intersecting z. The exact sequence above gives 0 -> O_l(-1) -> E_l -> O_l(1) -> 0 In particular, we can not say what E_l is. (b) if l is passing trough exactly one point in z, we get E_l(1) -> O_l(1) -> 0, or E_l -> O_l -> 0, which implies that E_l is isomorphic to O_l+O_l, and l is not a jumping line; (c) (b) if l is passing trough exactly two points in z, we get E_l -> O_l(-1) -> 0, i.e., E_l can not be O_l+O_l, in particular, l is a jumping line.